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Water equilibrium

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  1. Nov 20, 2014 #1
    Hi,
    I must calculate how much CO2 has to be removed from the water (pHs=7.2) to obtain equilibrium

    Given concentrations:
    HCO3-: 223 mg/L
    CO2: 60 mg/L
    pH: unknown
    SI: unknown

    What is meant by equilibrium? (ph=7?) Please advise how to tackle this problem
     
  2. jcsd
  3. Nov 20, 2014 #2

    Bystander

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    There are missing parameters: temperature, pressure, other solutes. Check the problem statement and see if isn't referring you to something previously mentioned.
     
  4. Nov 21, 2014 #3

    Borek

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    My bet is that they want you to assume concentration of CO2 is equivalent to the concentration of H2CO3. Then it is just a matter of treating the solution as a buffer, finding the required final concentration of carbonic acid, and comparing it with the initial (60 mg/L) CO2 concentration.
     
  5. Nov 21, 2014 #4
    Sorry, I forgot to mention that this is a water treatment problem. Temperature is 12C.
    What formula is applicable?
     
  6. Nov 21, 2014 #5

    Borek

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    Acid dissociation constant definition, or Henderson-Hasselbalch equation (which is actually the same, just rearranged to a more convenient form).
     
  7. Nov 22, 2014 #6
    Well, after some research I came up with this, what do you think?

    pHs = 7.2 = equilibrium of CO2 (= H2CO3 ) and HCO3-

    pHs = pK1 – log ([CO2] / [HCO3-])

    pK1 = 6.45 @12C

    log ([CO2] / [HCO3-]) = -0.750

    log ([CO2] / [HCO3-]) = 0.178

    [HCO3-] = 223/61 = 3.656 mol/L

    [CO2] = 0.178 * 3.656 = 0.650 mol/L

    CO2 = 0.650 * 44 = 28.6 mg/L

    We need to remove 60 – 28.6 = 31.4 mg/L
     
  8. Nov 22, 2014 #7

    Borek

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    Actually I did a mistake reading the original question, and it doesn't make sense to me now. Sorry about that. I missed the part about initial pH being 7.2. If teh initial pH is 7.2 and you are asked to change the pH of the solution to 7.0, you have to ADD acid, not remove it from the solution.

    However, if you use concentrations given to calculate pH (which I did) you will find that pH of the initial solution is around 6.8, not 7.2.

    Youdon't mean a log here, but OK.

    223 mg, not g. Besides, don't ignore units, as you force others to guess what is 61.
     
  9. Nov 23, 2014 #8
    Sorry for my typo's in the units. The question as I understand it is to obtain Carbonic acid equilibrium given the equilibrium pH = pHs = 7.2

    Then:

    pHs = 7.2 = equilibrium of CO2 (= H2CO3 ) and HCO3-

    pHs = pK1 – log ([CO2] / [HCO3-])

    pK1 = 6.45 @12C

    log ([CO2] / [HCO3-]) = -0.750

    [CO2] / [HCO3-] = 0.178

    [HCO3-] = 223/61 = 3.656 mmol/L

    [CO2] = 0.178 * 3.656 = 0.650 mmol/L

    CO2 = 0.650 * 44 = 28.6 mg/L

    We need to remove 60 – 28.6 = 31.4 mg/L. Do you think this is correct?
     
  10. Nov 23, 2014 #9

    Borek

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    At least logic behind looks OK, I have not checked the exact value (but it is definitely in a correct ballpark).
     
  11. Nov 30, 2014 #10
    Hi Borek,
    That calculation was correct!
    many thanks for your assistance.

    One more question, how do I calculate this one?:
    Lime (calcium hydroxide) is used in softening. How much lime (mmol/L) needs to be dosed in the softening step?
     
  12. Nov 30, 2014 #11

    Borek

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    Please start another thread (and follow the template).
     
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